12.1 Magnetic Fields

1. AIt is pulled toward the south pole, which acts like a negative charge.
   A student who treats magnetic poles as electric charges picks this. The south pole is not a negative charge; a magnet exerts no magnetic force on a charge at rest.
2. BThere is no magnetic force on it, since it is not moving at all. Correct
   A magnetic field exerts forces on moving charges, currents and magnetic materials. A charge at rest is none of these, so it feels no magnetic force, however strong the field.
3. CIt is pushed along B⃗ by a force of size qB, as in an electric field.
   A student who carries F⃗ = qE⃗ over to magnetic fields picks this. There is no magnetic force of the form qB along the field; a magnetic force on a moving charge is perpendicular to B⃗, and here the charge is not moving.
4. DIt is pushed at right angles to B⃗, as magnetic forces act sideways.
   A student who thinks every charge in a magnetic field feels a magnetic force picks this. Magnetic forces on charges are indeed sideways, but only on moving charges; this charge is at rest.

CED 12.1.A.1 · Read this in Fix

1. AB_P = B_Q > B_R
   A student who thinks a magnet's field depends only on distance picks this, because P and Q are on the same circle. The arrows at P and Q differ in length: at the same distance, the field on the axis is stronger than beside the magnet's middle.
2. BB_P = B_Q = B_R
   A student who pictures the field as a zone of uniform strength around the magnet picks this. The arrows have different lengths: the field's magnitude varies from point to point and decreases with distance.
3. CB_P > B_R > B_Q
   A student who thinks the middle of a magnet is unmagnetized, with almost no field beside it, puts Q last. The arrow at Q is longer than the one at R: beside the middle the field is weaker than on the axis at the same distance, but it is not close to zero.
4. DB_P > B_Q > B_R Correct
   In a vector field map, arrow length shows field magnitude. The arrow at P is longest, the one at Q is shorter and the one at R is shortest. P and Q are the same distance from the center, but a magnet's field is stronger on its axis than beside its middle.

CED 12.1.A.2 · Read this in Fix

1. APositive, as the north pole inside Z is a source of field
   A student who treats the north pole as a charge-like source picks this. There are no magnetic monopoles: the field lines that leave the north end entered Z through the body of the magnet.
2. BZero, as the flux leaving Z equals the flux entering it Correct
   Gauss's law for magnetism, ∮B⃗·dA⃗ = 0, holds for every closed surface. Inside the magnet the field points from S toward N, so field enters Z through the part of the surface crossed by the magnet; the same field lines then leave through the rest of Z and loop back. Inward and outward flux cancel.
3. CNegative, as the field points into the north end of the magnet
   A student who reverses the pole convention picks this. Outside the magnet the field points away from the north end; in any case, the net flux through a closed surface is zero.
4. DUnknown, as Z is not symmetric and B varies over it
   A student who thinks Gauss's law needs symmetry picks this. ∮B⃗·dA⃗ = 0 holds for every closed surface; symmetry is needed only to find the field itself.

CED 12.1.A.3.i · Read this in Fix

1. Aqv/r
   A student who takes the number of revolutions per second to be v/r, the angular speed ω, picks this. The number of revolutions per second is ω/(2π) = v/(2πr).
2. Bq
   A student who thinks current is the amount of charge moving picks this. Current is a rate (charge per unit time), so it must depend on how often the charge goes round; q has units of C, not A.
3. Cqv/(2πr) Correct
   Current is the rate at which charge passes a point, I = dq/dt. The particle goes round once every T = 2πr/v, so charge q passes each point once per period: I = q/T = qv/(2πr).
4. D2πrq/v
   A student who multiplies the charge by the period, 2πr/v, instead of dividing by it picks this. Current is charge per unit time, I = q/T = qv/(2πr); 2πrq/v has units of C·s, not A.

Working The particle passes any point on the circle once per period, T = 2πr/v. Current is charge per unit time: I = q/T = qv/(2πr). Distractors: revolutions per second taken as v/r (ω) → qv/r; current taken as the amount of charge → q; charge multiplied by the period instead of divided by it → q(2πr/v) = 2πrq/v.

CED 12.1.B.1 · Read this in Fix

1. AParallel to the magnet, toward its S end Correct
   A compass needle is a magnetic dipole and aligns with the field at its location, its north end pointing along the field. Outside the magnet the field runs from the N end round to the S end; directly above the middle it is parallel to the magnet, pointing toward the S end.
2. BAlong a straight line from P to the magnet's S end
   A student who thinks a compass points straight at the attracting pole picks this. The needle lines up with the field line through P, which is parallel to the magnet there, not aimed at the pole.
3. CParallel to the magnet, toward the magnet's N end
   A student who thinks a compass's north end is drawn to a north pole picks this. The needle's north end is a north pole, repelled by the magnet's N end; it points along the field, toward the S end.
4. DStraight up, directly away from the magnet
   A student who pictures the field leaving the magnet at right angles, like the field of a charged conductor, picks this. Beside the magnet's middle the field lines run parallel to the magnet.

CED 12.1.B.2 · Read this in Fix

1. AIron is denser than aluminum, so there is more metal for the magnet to pull on.
   A student who thinks a magnet pulls harder on more massive objects picks this. The difference is in the type of material: a much heavier block of aluminum would still barely respond.
2. BAluminum is not a magnetic material, so the field exerts no force on it.
   A student who thinks 'non-magnetic' means 'unaffected' picks this. Aluminum is weakly attracted (paramagnetic); the force exists but is too small to see with an ordinary magnet.
3. CIron's atomic dipoles align strongly with the field; aluminum's align only weakly. Correct
   A material's composition sets its magnetic behavior. Iron is ferromagnetic: its dipoles align strongly with an external field, so it is strongly attracted. Aluminum is paramagnetic: its dipoles align only weakly, so the attraction is too small to notice.
4. DThe field creates dipoles in both rods, but it creates far more of them in the iron.
   A student who thinks an external field creates the dipoles in a material picks this. The atoms of both metals already have dipoles; the field only aligns them, strongly in iron and weakly in aluminum.

CED 12.1.B.3 · Read this in Fix

1. ASouthward and upward, out of the ground
   A student who takes the magnetic pole near geographic north to be a north magnetic pole, with the field pointing out of it, picks this. The compass at E shows the field pointing north, toward that pole, so the pole there is a south magnetic pole and the field at P points north and down.
2. BNorthward only, parallel to the ground
   A student who thinks Earth's field is horizontal everywhere picks this. It is roughly horizontal only near the equator; in the northern hemisphere the dipole field also points down into the ground.
3. CStraight up, away from the center of Earth
   A student who pictures a magnet's field leaving its surface at right angles picks this. A dipole's field lines curve from one hemisphere to the other; at P the field points north and down.
4. DNorthward and also downward, into the ground Correct
   The compass at E shows that the field points north at the equator, so the field lines run from the southern hemisphere to the northern one: the dipole's south pole is near geographic north. In a dipole's field, lines leave the ground in the southern hemisphere and re-enter it in the northern hemisphere, so at P the field points north and down.

CED 12.1.B.4 · Read this in Fix

1. AX has more space between its atoms, so field lines can pass through it.
   A student who reads 'permeable' in its everyday sense (letting something soak through) picks this. Permeability measures magnetization in response to a field, not porosity.
2. BX keeps its magnetization after the field is removed, but Y does not.
   A student who equates high permeability with permanent magnetism picks this. Permeability describes the response while the field acts; whether magnetization remains afterward is a separate property.
3. CIn that field and at that temperature, X becomes more magnetized than Y. Correct
   Magnetic permeability is a measure of how much a material becomes magnetized in response to an external field. Under the conditions of the comparison, the larger permeability means the stronger magnetization.
4. DX has the larger permeability at any temperature and in any field.
   A student who treats permeability as a fixed property of each material, like density, picks this. Permeability varies with temperature and field strength, so one comparison does not show that X's permeability is larger under every condition.

CED 12.1.C.1 · Read this in Fix

1. AThe value is plausible: μ₀ belongs to vacuum, and air is matter. Correct
   μ₀ is the constant permeability of free space. The permeability of matter differs from it because of the matter's composition; air's molecules respond weakly to a magnetic field, so its permeability is very slightly different from μ₀.
2. BThe student is right: μ₀ is defined as the permeability of air.
   A student who remembers the approximation 'air ≈ vacuum' as a definition picks this. μ₀ is defined for free space; air's permeability is only extremely close to it.
3. CThe value is plausible, since μ₀ varies slightly with the weather.
   A student who thinks μ₀ changes with conditions picks this. μ₀ is a constant; it is air's permeability, not μ₀, that can differ from μ₀.
4. DThe student is right: gas molecules have no magnetic dipoles at all.
   A student who thinks gases are magnetically 'empty' picks this. All matter is at least weakly diamagnetic, and oxygen molecules are paramagnetic, so air responds slightly to a magnetic field.

CED 12.1.C.2 · Read this in Fix

1. AOne value must be wrong, since a material has only a single permeability.
   A student who treats permeability as a fixed property of the material picks this. The same core can have different permeabilities under different conditions.
2. BBoth values should be μ₀, since permeability is a universal constant.
   A student who thinks every material's permeability is μ₀ picks this. μ₀ is the permeability of free space; nickel, a ferromagnetic material, has a permeability very different from μ₀.
3. CBoth values can be correct, since permeability varies with temperature. Correct
   The permeability of matter is not a constant for a material: it varies with factors including temperature, orientation and the strength of the external field. Different values at different temperatures are expected.
4. DThe values differ because μ₀ changes as the temperature rises.
   A student who thinks μ₀ depends on conditions picks this. μ₀ is a constant; what changes with temperature is the nickel's own permeability.

CED 12.1.C.3 · Read this in Fix

1. AA small charged bead held at rest by the box is pushed sideways, away from it.
   A student who thinks a magnetic field pushes on any charge picks this. A charge at rest feels no magnetic force, so a push on the bead would point to an electric field, not a magnetic one.
2. BA copper coin lying at rest next to the box is pulled strongly toward it.
   A student who thinks magnets attract all metals picks this. Copper interacts only very weakly with a magnetic field (it is weakly repelled), so strong attraction of a copper coin is not what a magnet produces.
3. CThe leaves of an uncharged electroscope spread apart when it is brought near the box.
   A student who treats magnetism as a kind of static electricity picks this. Electroscope leaves spread because of electric charge; a magnet's field does not charge an electroscope, so this would not show a magnetic field.
4. DA compass needle carried around the box points different ways at different places. Correct
   A magnetic field is a vector field that determines the magnetic force on magnetic materials, including a compass needle, which aligns with it. Earth's field alone would make the needle point the same way everywhere in the room; a changing direction around the box is evidence of a field from the box.

CED 12.1.A.1 · Read this in Fix

1. AEvery source is a dipole or a combination of dipoles; none is a single pole. Correct
   Magnetic fields are produced by magnetic dipoles or combinations of dipoles, never by monopoles. A bar magnet, a compass needle and a current loop are all dipoles, each with both a north and a south pole.
2. BEach end of a bar magnet is a separate source, like a single point charge.
   A student who pictures each pole as a charge-like source picks this. The two poles of a magnet are not separate sources: the field lines loop through the magnet, and no isolated pole has ever been found.
3. COnly permanent magnets are sources; moving charges make electric fields only.
   A student who separates magnetism from electricity picks this. Moving charges do produce magnetic fields: magnetic dipoles result from the circular or rotational motion of charges, including electrons in magnets.
4. DA charged object at rest is a source, with its charge acting as a magnetic pole.
   A student who treats poles as electric charges picks this. A charge at rest produces an electric field but no magnetic field; charge is not a magnetic pole.

CED 12.1.A.1.i · Read this in Fix

1. AIt is a single north pole, as all of the charge on it is positive.
   A student who equates positive charge with a north pole picks this. The sign of the charge fixes which side is north, not whether both poles exist; the sphere is a dipole with north and south polarity.
2. BIt produces an electric field only, and no magnetic field at all.
   A student who thinks charges make only electric fields picks this. The charge is moving, and the rotational motion of charge is exactly what produces a magnetic dipole.
3. CIt has north and south poles, at opposite ends of its spin axis. Correct
   The charge moves in circles about the spin axis, and circular motion of charge produces a magnetic dipole. Every dipole has north and south polarity; for this sphere the two poles lie on opposite sides, at the ends of the spin axis.
4. DIt is magnetic, but a sphere has no ends at which its two poles could form.
   A student who links poles to the ends of long magnets picks this. Polarity belongs to every dipole, whatever its shape: the external field points away from one side of the sphere (north) and returns to the opposite side (south).

CED 12.1.A.1.ii · Read this in Fix

1. A5.0 × 10⁻⁴ T Correct
   The two fields are vectors at right angles, so the net field is found with the Pythagorean theorem: √((3.0 × 10⁻⁴ T)² + (4.0 × 10⁻⁴ T)²) = 5.0 × 10⁻⁴ T.
2. B7.0 × 10⁻⁴ T
   A student who adds the field magnitudes picks this: 3.0 × 10⁻⁴ T + 4.0 × 10⁻⁴ T. That would be correct only if the fields pointed the same way; here they are perpendicular.
3. C4.0 × 10⁻⁴ T
   A student who keeps only the stronger field picks this. Magnet 1's field is also present at P and adds to magnet 2's as a vector.
4. D1.0 × 10⁻⁴ T
   A student who thinks the fields of two north poles facing P must oppose each other subtracts them: 4.0 × 10⁻⁴ T − 3.0 × 10⁻⁴ T. The figure shows the two fields at right angles, not opposite.

Working Magnetic fields add as vectors. B₁ (3.0 × 10⁻⁴ T, to the right) and B₂ (4.0 × 10⁻⁴ T, up) are perpendicular, so B = √(B₁² + B₂²) = √((3.0)² + (4.0)²) × 10⁻⁴ T = 5.0 × 10⁻⁴ T, directed up and to the right, at 53° above the rightward direction.

CED 12.1.A.2 · Read this in Fix

1. ARegion 1 has more, since the field inside the magnet is stronger.
   A student who treats flux as field strength picks this. The field inside the magnet is stronger, but region 1 is small; the weaker field outside spreads over a much larger area, and the fluxes balance.
2. BRegion 2 has more, since its area is far larger than that of region 1.
   A student who thinks a larger surface always has more flux picks this. The outside field is much weaker than the field inside the magnet, and every loop that crosses region 2 has also crossed region 1.
3. CThey are equal, as each field line crosses region 1 once and region 2 once. Correct
   Magnetic field lines form closed loops (∮B⃗·dA⃗ = 0). Each loop passes through the magnet from S to N, crossing region 1, and returns outside from N to S, crossing region 2 in the opposite direction. So the flux through region 1 and the flux through region 2 are equal in magnitude and opposite in sign.
4. DRegion 1 has none, since the field lines stop at the magnet's poles.
   A student who thinks field lines begin and end at the poles picks this. The lines continue through the magnet from S to N, so there is a strong field, and a large flux, in region 1.

Working Take the plane together with a very large hemisphere far away (where the dipole field is negligible) as a closed surface: ∮B⃗·dA⃗ = 0, so the flux through the whole plane is zero. Inside the magnet (region 1) the field points from S to N; outside (region 2) it points from N back toward S. So Φ₁ = −Φ₂ and |Φ₁| = |Φ₂|: every closed field line crosses region 1 once and region 2 once, in opposite directions.

CED 12.1.A.3 · Read this in Fix

1. A7.5 mWb
   A student who multiplies the field by the full area of the sloping face, (0.15 T)(0.50 m)(0.10 m), picks this. The sloping face is tilted, so only the field component perpendicular to it counts.
2. B6.0 mWb
   A student who uses the cosine of the angle between the field and the sloping face itself (cos 37° = 0.80) picks this: (0.15 T)(0.050 m²)(0.80). The angle in the flux equation is measured from the face's area vector, which gives cos θ = 0.60.
3. C4.5 mWb Correct
   The net flux through the closed wedge is zero. Only the vertical face and the sloping face are crossed by the field; the field enters the vertical face with flux (0.15 T)(0.30 m)(0.10 m) = 4.5 mWb, so the same 4.5 mWb must leave through the sloping face.
4. D0.0 mWb
   A student who reads Gauss's law for magnetism as 'zero flux through every face' picks this. It is the net flux through the closed wedge that is zero: 4.5 mWb enters through the vertical face and 4.5 mWb leaves through the sloping face.

Working Gauss's law for magnetism: the net flux through the closed wedge is zero. The field is parallel to the bottom face and to the two triangular end faces, so their flux is zero. The field enters the vertical face (area 0.30 m × 0.10 m = 0.030 m²) head-on: Φ = −BA = −(0.15 T)(0.030 m²) = −4.5 × 10⁻³ Wb. So the sloping face carries +4.5 × 10⁻³ Wb = 4.5 mWb. (Check directly: sloping face area (0.50 m)(0.10 m) = 0.050 m²; its area vector makes an angle θ with B⃗ where cos θ = 0.30/0.50 = 0.60; Φ = (0.15 T)(0.050 m²)(0.60) = 4.5 mWb.)

CED 12.1.A.3.i · Read this in Fix

1. A0
   A student who takes the field to be purely axial picks this. Because B_z doubles along the cylinder, more flux leaves the top than enters the bottom; ∮B⃗·dA⃗ = 0 then requires flux through the side, so the field must have a radial component there.
2. BB₀R/d
   A student who balances the side against the top face only, leaving out the bottom face, picks this: |Φ_side| = 2πR²B₀, divided by 2πRd. The bottom face is part of the closed surface and carries −πR²B₀.
3. CB₀
   A student who equates the radial field to the change in the axial field along the cylinder, B₀, picks this, balancing fields instead of fluxes. The flux balance involves the areas: an end area πR² against a side area 2πRd.
4. DB₀R/(2d) Correct
   The net flux through the closed cylinder is zero. The end faces give 2πR²B₀ − πR²B₀ = πR²B₀ net outward, so the side must carry −πR²B₀. Dividing by the side's area, 2πRd, gives a radial component of magnitude B₀R/(2d), directed toward the axis.

Working ∮B⃗·dA⃗ = 0 over the closed cylinder. Top end (z = d, area vector +z): Φ_top = πR²B₀(1 + 1) = 2πR²B₀. Bottom end (z = 0, area vector −z): Φ_bottom = −πR²B₀. So Φ_side = −(Φ_top + Φ_bottom) = −πR²B₀. On the side, Φ_side = B_r(2πRd), so B_r = −B₀R/(2d): magnitude B₀R/(2d), pointing toward the axis. Distractors: field purely axial → 0; bottom face omitted → |Φ_side| = 2πR²B₀, giving B₀R/d; change in axial field over the cylinder (B₀) equated to the radial field → B₀.

CED 12.1.A.3.i · Read this in Fix

1. AP: toward the S end; Q: toward the S end
   A student who applies 'field lines go from N to S' inside the magnet too picks this. The rule describes the external field only; inside, the closed loops return from S to N.
2. BP: toward the N end; Q: toward the S end Correct
   Outside a bar magnet the field points away from the north end and loops round to the south end, so beside the middle it runs from the N end toward the S end. Field lines form closed loops, so inside the magnet the field completes each loop, pointing from the S end toward the N end.
3. CP: toward the S end; Q: toward the N end
   A student who thinks the external field points into the north end reverses both directions. By definition the external field points away from the north pole, so at Q it points toward the S end.
4. DP: zero field there; Q: toward the S end
   A student who thinks field lines start and stop at the poles picks this. The lines continue through the magnet, where the field is strong and points from S to N.

CED 12.1.A.3.ii · Read this in Fix

1. AQωr dr/(πR²) Correct
   The charge is spread over the disk's area, σ = Q/(πR²). The ring between r and r + dr has area 2πr dr, so dq = 2Qr dr/R². The ring goes round ω/(2π) times per second, so this charge passes a fixed point ω/(2π) times per second: dI = dq·ω/(2π) = Qωr dr/(πR²).
2. B2Qωr dr/R²
   A student who takes the number of passes per second to be ω instead of ω/(2π) picks this: dI = dq·ω, which is 2π times too large. The ring goes round, and its charge passes a fixed point, ω/(2π) times per second.
3. CQω dr/(2πR)
   A student who spreads the charge evenly along the radius, dq = (Q/R)dr, picks this: dI = (Q/R)dr·ω/(2π). The charge is spread over the area, so a ring's charge is proportional to its circumference: dq = 2Qr dr/R².
4. D2Qr dr/R²
   A student who takes the ring's charge itself as its current, dI = dq, picks this. Current is the rate at which charge passes a point, so it must depend on how fast the disk spins; 2Qr dr/R² has units of C, not A.

Working Charge per unit area: σ = Q/(πR²). The thin ring between r and r + dr has area 2πr dr, so dq = σ(2πr dr) = 2Qr dr/R². The ring turns ω/(2π) times per second, so its charge passes a fixed point ω/(2π) times per second: dI = dq·ω/(2π) = Qωr dr/(πR²). Distractors: passes per second taken as ω → dI = dq·ω = 2Qωr dr/R²; charge spread evenly along the radius, dq = (Q/R)dr → dI = Qω dr/(2πR); current taken as the ring's charge itself → dI = dq = 2Qr dr/R², which is in C, not A.

CED 12.1.B.1 · Read this in Fix

1. AΦ₃ > Φ₁ = Φ₂
   A student who treats the enclosed north pole as a source of extra flux picks this. The field lines that leave the north end inside surface 3 entered it through the body of the magnet, so enclosing the pole adds no net flux.
2. BΦ₁ = Φ₂ = Φ₃ Correct
   Surfaces 1 and 3 together make a closed surface around the magnet's north end, and the net flux through any closed surface is zero, because there are no magnetic monopoles. So the flux through 3 equals the flux through 1 (both oriented to the right), and the same argument applies to surfaces 1 and 2.
3. CΦ₃ > Φ₂ > Φ₁
   A student who thinks a larger surface catches more flux ranks the surfaces by size. Surfaces with the same boundary carry the same magnetic flux, however large they are.
4. DΦ₁ > Φ₂ > Φ₃
   A student who thinks that curved surfaces, tilted away from the field, catch less flux than the flat one picks this. The extra area of a curved surface makes up exactly for the tilt, and field that crosses a surface twice cancels itself.

Working Any two of the surfaces together form a closed surface, and ∮B⃗·dA⃗ = 0 for every closed surface, including 1 + 3, which encloses the magnet's north end. Reversing the area vectors of one surface to make them all point outward, the fluxes through the two surfaces must be equal when both are oriented to the right. So Φ₁ = Φ₂ = Φ₃.

CED 12.1.A.1.i · Read this in Fix

1. AEach pole acts like an electric charge, and either sign of charge attracts a neutral nail.
   A student who treats magnetic poles as electric charges picks this: a charge of either sign attracts a neutral conductor by induction, so both poles would seem to attract the nail. Magnetic poles are not charges, and a magnet exerts no electric force on the nail; the attraction comes from magnetism induced in the iron.
2. BMagnets attract every metal object, whatever the metal happens to be made of.
   A student who thinks all metals are attracted by magnets picks this. Most metals, such as copper and aluminum, show no noticeable attraction; iron is attracted because it is ferromagnetic.
3. CMagnetism flows out of the magnet and into the nail, pulling the nail along with it.
   A student who pictures magnetism as a substance passed from the magnet to the nail picks this. Nothing flows into the nail: the magnet's field aligns the dipoles the nail already has.
4. DA pole's field aligns dipoles in the nail, so its near end becomes an unlike pole. Correct
   The nail's atomic dipoles (in domains) point in many directions, so it starts unmagnetized. A pole's field aligns them, inducing magnetism: the end near a north pole becomes a south pole and the end near a south pole becomes a north pole. Unlike poles attract, so either pole attracts the nail.

CED 12.1.B.1.i · Read this in Fix

1. AOne end of the rod repels the north end of a compass needle. Correct
   Unmagnetized steel is attracted by either pole of a compass needle, because the needle's field induces magnetism in it; it is never repelled. Repulsion between the rod's end and the needle's north end shows that the end is a north pole that persists without the magnet: the rod's dipoles have stayed aligned.
2. BEach end of the rod attracts the north end of a compass needle.
   A student who takes attraction as proof of magnetism picks this. An unmagnetized steel rod would do exactly this, since the needle induces magnetism in it; only repulsion shows permanent magnetism.
3. CThe rod picks up small bits of paper after it is rubbed with a cloth.
   A student who treats magnetism as static electricity picks this. Picking up paper after rubbing shows electric charge, which says nothing about magnetization.
4. DThe magnet used for stroking is now weaker than it was before.
   A student who thinks stroking passes magnetism from the magnet into the rod picks this. Magnetizing the rod aligns its own dipoles; it does not drain the magnet, so a weaker magnet would not be evidence for the claim.

CED 12.1.B.1.i · Read this in Fix

1. AX stays a north pole and Y stays a south pole.
   A student who thinks the magnet has a 'north half' and a 'south half' picks this, taking the whole left piece to stay north and the whole right piece to stay south. Each piece is a complete dipole with one north and one south end.
2. BX and Y are not poles; each piece has one pole.
   A student who thinks a north pole can exist alone picks this. No magnetic north pole is ever found in isolation: each half is a dipole, with new poles at X and Y.
3. CNeither of the pieces is a magnet after the break.
   A student who thinks a magnet only works as a whole picks this. Breaking the magnet does not undo the alignment of dipoles inside each piece, so both pieces are magnets.
4. DX is now a south pole and Y is now a north pole. Correct
   Each piece is still a magnetic dipole, because the aligned dipoles in each piece keep pointing the same way, from the S end toward the N end of the original magnet. So the left piece has N at its left end and S at X, and the right piece has N at Y and S at its right end.

CED 12.1.B.1.ii · Read this in Fix

1. A3.0 N
   A student who thinks the floating magnet, not touching anything below it, exerts no force on the lower magnet picks this: only (0.30 kg)(10 m/s²). The magnets interact at a distance, and the force on the lower magnet is equal and opposite to the force on the upper one.
2. B2.4 N
   A student who thinks the repulsion pushes the lower magnet upward too picks this: (0.30 kg − 0.060 kg)(10 m/s²). Repulsion pushes the two magnets apart, so the force on the lower magnet points down.
3. C3.6 N Correct
   The floating magnet is repelled upward with a force equal to its weight, (0.060 kg)(10 m/s²) = 0.60 N. Repulsion pushes the magnets apart, so the lower magnet is pushed down with the same 0.60 N (Newton's third law). The scale supports the lower magnet's weight, 3.0 N, plus this force: 3.6 N.
4. D4.2 N
   A student who adds both the upper magnet's weight and the magnetic force to the lower magnet's weight picks this: (0.30 kg + 2 × 0.060 kg)(10 m/s²). The upper magnet's weight acts on the upper magnet; the lower magnet feels only the magnetic force, which equals that weight in size.

Working Like poles face each other, so the magnets repel; the upper magnet rests in equilibrium without touching. Upper magnet in equilibrium (smooth rod, no vertical friction): magnetic force up = its weight = (0.060 kg)(10 m/s²) = 0.60 N. By Newton's third law the upper magnet pushes the lower one down with 0.60 N. Lower magnet: scale force = its weight + 0.60 N = (0.30 kg)(10 m/s²) + 0.60 N = 3.0 N + 0.60 N = 3.6 N.

CED 12.1.B.1.iii · Read this in Fix

1. A1.8 × 10⁻⁵ T
   A student who assumes the inverse-square law of a point charge picks this: (7.2 × 10⁻⁵ T)/4. The data show a factor of 8, not 4, for each doubling of distance.
2. B9.0 × 10⁻⁶ T Correct
   The data show that doubling the distance divides the field by about 8. Going from 0.20 m to 0.40 m is another doubling, so the field is divided by 8 again: (7.2 × 10⁻⁵ T)/8 = 9.0 × 10⁻⁶ T.
3. C3.6 × 10⁻⁵ T
   A student who assumes that twice the distance gives half the field picks this: (7.2 × 10⁻⁵ T)/2. The measurements show a much faster decrease.
4. D1.1 × 10⁻⁶ T
   A student who treats each extra 0.10 m as another factor of 8 picks this: two more steps, (7.2 × 10⁻⁵ T)/64. The factor of 8 belongs to each doubling of distance, and 0.20 m to 0.40 m is one doubling.

Working Doubling the distance from 0.10 m to 0.20 m divides the field by (5.8 × 10⁻⁴ T)/(7.2 × 10⁻⁵ T) ≈ 8 = 2³, so B ∝ 1/r³. Doubling again, from 0.20 m to 0.40 m, divides it by 8 again: B = (7.2 × 10⁻⁵ T)/8 = 9.0 × 10⁻⁶ T.

CED 12.1.B.1.iv · Read this in Fix

1. AX and Y could both be iron, as all metals act alike.
   A student who thinks all metals behave like iron picks this. The graphs differ by a factor of about a million in B_s, and only X stays magnetized; Y cannot be iron.
2. BX could be iron, while Y could be aluminum. Correct
   X responds very strongly along the field and keeps a large B_s after t₂: its dipoles stay aligned, as in a ferromagnetic material such as iron. Y responds about a million times more weakly, along the field, and loses its alignment when the field is removed, as a paramagnetic material such as aluminum does.
3. CY could be copper, as diamagnets align along a field.
   A student who thinks diamagnetic dipoles align along the field picks this. Diamagnetic materials such as copper align weakly opposite the field, which would make B_s negative; Y's B_s is positive.
4. DX could be copper, as good conductors respond most.
   A student who links magnetic response to electrical conductivity picks this. Copper conducts very well but is diamagnetic: its response would be tiny and opposite the field, not large and lasting like X's.

CED 12.1.B.3.i · Read this in Fix

1. AIts dipoles stay aligned, so it is left as a weak permanent magnet.
   A student who pictures a paramagnet as a weak ferromagnet picks this. Paramagnetic materials do not keep their alignment after the external field is removed.
2. BNothing changed at any stage, as aluminum does not respond to fields at all.
   A student who thinks 'non-magnetic' materials are unaffected picks this. Aluminum's dipoles did align weakly while the field was on; the effect is small, not absent.
3. CIts dipoles, which the field created, vanish once the field is gone.
   A student who thinks the field creates the dipoles picks this. Aluminum's atoms have dipoles all the time; the field only aligned them, and after it is removed they are still there, pointing randomly.
4. DIts dipoles return to random directions, so it is no longer magnetized. Correct
   Aluminum is paramagnetic. While the field is on, its atomic dipoles align weakly with the field; the alignment does not remain once the field is removed, so the rod is left unmagnetized.

CED 12.1.B.3.ii · Read this in Fix

1. AF_Fe > F_Al > F_Cu Correct
   Iron is ferromagnetic and strongly attracted, so F_Fe is large and positive. Aluminum is paramagnetic: weakly attracted, small positive F_Al. All materials are diamagnetic, and in copper this is the main effect: its dipoles align weakly opposite the field, so it is weakly repelled and F_Cu is negative.
2. BF_Fe > F_Al = F_Cu
   A student who thinks non-ferromagnetic materials feel no force puts aluminum and copper equal at zero. Aluminum is weakly attracted and copper weakly repelled.
3. CF_Fe > F_Cu > F_Al
   A student who swaps paramagnetism and diamagnetism thinks copper is weakly attracted and aluminum weakly repelled, and picks this. Aluminum (paramagnetic) aligns weakly along the field and is attracted; copper (diamagnetic) aligns weakly opposite it and is repelled.
4. DF_Fe = F_Al = F_Cu
   A student who thinks all metals are attracted equally by magnets picks this. Iron is attracted far more strongly than aluminum, and copper is not attracted at all but weakly repelled.

Working Iron is ferromagnetic: strongly attracted, large positive F. Aluminum is paramagnetic: dipoles align weakly along the field, weakly attracted, small positive F. Copper is diamagnetic: dipoles align weakly opposite the field, weakly repelled, small negative F. So F_Fe > F_Al > F_Cu (with F_Cu < 0).

CED 12.1.B.3.iii · Read this in Fix

1. AIt feels no force, since water contains no magnetic material of any kind.
   A student who thinks 'non-magnetic' means 'unaffected' picks this. Diamagnetism is a property of all materials; in a very strong field the weak repulsion of water can be seen.
2. BIt is weakly repelled, as its dipoles align weakly opposite the field. Correct
   All materials are diamagnetic: their electronic structure produces a usually weak alignment of dipole moments opposite the external field. Water has no stronger magnetic behavior to mask this, so it is weakly pushed away from the pole.
3. CIt is weakly attracted, as its dipoles align weakly along the field.
   A student who mixes up diamagnetism and paramagnetism picks this. Water is diamagnetic: its dipole moments align opposite the field, so it is repelled.
4. DAny force on it comes from electric charge that it may carry.
   A student who ties magnetic forces to charge picks this. A drop at rest feels no magnetic force from its charge; the force it does feel comes from its diamagnetism, charged or not.

CED 12.1.B.3.iii · Read this in Fix

1. Aμ_Fe > μ_Al = μ₀ = μ_Cu
   A student who thinks materials that are not ferromagnetic are unaffected by fields gives them the permeability of free space. Aluminum's weak paramagnetism and copper's weak diamagnetism make both differ slightly from μ₀.
2. Bμ_Fe > μ_Al > μ₀ > μ_Cu Correct
   The permeability of matter differs from μ₀ because of its composition. Iron's dipoles align strongly (μ_Fe ≫ μ₀); aluminum's align weakly along the field (μ_Al a little above μ₀); copper's diamagnetic response is weakly opposite the field (μ_Cu a little below μ₀).
3. Cμ_Fe > μ_Cu > μ₀ > μ_Al
   A student who swaps paramagnetism and diamagnetism puts copper above μ₀ and aluminum below it. Aluminum (paramagnetic) aligns along the field, raising μ; copper (diamagnetic) aligns against it, lowering μ.
4. Dμ_Cu > μ_Al > μ_Fe > μ₀
   A student who links magnetic response to electrical conductivity ranks the metals by conductivity (copper, then aluminum, then iron). Permeability depends on how a material's dipoles respond, and iron's is by far the largest.

Working Iron is ferromagnetic: μ_Fe is far larger than μ₀. Aluminum is paramagnetic (weak alignment along the field): μ_Al slightly larger than μ₀. Copper is diamagnetic (weak alignment opposite the field): μ_Cu slightly smaller than μ₀. So μ_Fe > μ_Al > μ₀ > μ_Cu.

CED 12.1.C.3 · Read this in Fix

1. AIron has one fixed permeability, and the curve just shows measuring errors.
   A student who treats permeability as a fixed material constant, like density, picks this. The smooth, large change with field strength is real: permeability varies with the field.
2. BThe permeability of iron keeps increasing as the external field grows.
   A student who expects 'more field, more permeability' picks this without following the curve. The permeability peaks near 0.3 mT and then falls.
3. CThe peak marks the field at which iron becomes a permanent magnet.
   A student who equates permeability with permanent magnetization picks this. The graph shows the response while the field acts; it says nothing about what remains when the field is removed.
4. DThe permeability of iron depends on the field, so it is not a constant. Correct
   The permeability of the same sample changes with the strength of the external field: it rises to a maximum and then falls. Permeability is not a constant for a material; it depends on conditions such as field strength and temperature.

Working Reading the graph: μ ≈ 1.2 × 10⁻³ T·m/A at zero field, a maximum of about 6 × 10⁻³ T·m/A near 0.3 mT, and about 1.4 × 10⁻³ T·m/A at 2.0 mT. The permeability changes by a factor of about 5 with field strength, so it is not a constant of the material.

CED 12.1.C.3 · Read this in Fix

1. Ae²/(4π√(πε₀mr³)) Correct
   The electric force is the centripetal force: e²/(4πε₀r²) = mv²/r, so v = e/(2√(πε₀mr)). The electron passes any point v/(2πr) times per second, so I = ev/(2πr). Substituting v gives e²/(4π√(πε₀mr³)).
2. Be²/(2π√(2πε₀mr³))
   A student who finds the speed by setting the kinetic energy equal to the magnitude of the electric potential energy, ½mv² = e²/(4πε₀r), gets v = e/√(2πε₀mr), which is √2 times too large, and so this current. That condition describes an electron with just enough energy to escape; a circular orbit requires e²/(4πε₀r²) = mv²/r.
3. Ce²/(2√(πε₀mr³))
   A student who takes the current as eω = ev/r instead of ev/(2πr) picks this, a value 2π times too large. The electron passes a given point once per revolution, v/(2πr) times per second.
4. De²/(2√(πε₀mr))
   A student who takes the current as the charge times its speed, I = ev, picks this. Current is the charge passing a point per second, ev/(2πr); e²/(2√(πε₀mr)) has units of C·m/s, not A.

Working Newton's second law, radial: e²/(4πε₀r²) = mv²/r, so v = e/√(4πε₀mr) = e/(2√(πε₀mr)). The electron passes a fixed point v/(2πr) times per second, so I = ev/(2πr) = e²/(4π√(πε₀mr³)) (equivalently e²√(k/(mr³))/(2π) with k = 1/(4πε₀)). Distractors: ½mv² = e²/(4πε₀r) gives v = e/√(2πε₀mr) and I = e²/(2π√(2πε₀mr³)); I = eω = ev/r gives e²/(2√(πε₀mr³)); I = ev gives e²/(2√(πε₀mr)), which is in C·m/s, not A.

CED 12.1.B.1 · Read this in Fix

1. AπB₀R²
   A student who multiplies the disk's area by the field on the axis, B₀, picks this, as if the field were uniform. B_z falls to B₀/2 at the rim, so each ring must be counted with its own field, and the flux is less than πB₀R².
2. B(5/6)B₀R
   A student who integrates the field along one radius, ∫₀ᴿ B_z dr, picks this, leaving out the ring area 2πr dr. The result is not even a flux: its unit is T·m, not T·m².
3. C0
   A student who applies 'zero flux' to each part of a closed surface picks this. ∮B⃗·dA⃗ = 0 holds for the closed surface made of the hemisphere and the disk as a whole: the flux leaving through the hemisphere equals the flux entering through the disk, and neither is zero.
4. D(3/4)πB₀R² Correct
   The hemisphere and the disk of radius R in the plane z = 0 form a closed surface, so ∮B⃗·dA⃗ = 0 makes the flux up through the hemisphere equal to the flux up through the disk. Summing over thin rings of area 2πr dr: Φ = ∫₀ᴿ B₀(1 − r²/(2R²))2πr dr = πB₀R² − πB₀R²/4 = (3/4)πB₀R².

Working The hemisphere and the flat disk of radius R in the plane z = 0 together form a closed surface, so by ∮B⃗·dA⃗ = 0 the flux through the hemisphere equals the flux through the disk (both counted in the +z sense). Only B_z crosses the disk. With rings of radius r and area 2πr dr: Φ = ∫₀ᴿ B₀(1 − r²/(2R²))2πr dr = 2πB₀(R²/2 − R²/8) = (3/4)πB₀R². Distractors: field taken as B₀ over the whole disk → πB₀R²; area element dr → ∫₀ᴿ B_z dr = B₀(R − R/6) = (5/6)B₀R, in T·m; zero flux applied to part of a closed surface → 0.

CED 12.1.A.3.i · Read this in Fix